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Math Help - Rings and Ideals

  1. #1
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    Rings and Ideals

    Need some help...

    Let \mathbb{Z}[\sqrt{-5}] the ring of gaussian integers.

    1)Show that <3>=<3;1+\sqrt{-5}><3;1-\sqrt{-5}>
    ( <a> means the ideal generated by a)

    2)Prove that the ideal on the right is prime.

    Thank you folks!
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  2. #2
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    Quote Originally Posted by Inti View Post

    Let \mathbb{Z}[\sqrt{-5}] the ring of gaussian integers.

    1) Show that <3>=<3,1+\sqrt{-5}><3,1-\sqrt{-5}>
    Hint: we have 3=3^2 - (1+\sqrt{-5})(1-\sqrt{-5}). thus: <3> \subseteq <3,1+\sqrt{-5}><3,1-\sqrt{-5}>. for the other side of the inclusion, note that (1+\sqrt{-5})(1-\sqrt{-5})=6 \in <3>.



    2) Prove that the ideals on the right is prime.
    Hint: define the map f: \mathbb{Z}[\sqrt{-5}] \longrightarrow \mathbb{Z}_3 by f(a+b\sqrt{-5})=a-b. show that f is a well-defined surjective ring homomorphism and: \ker f = <3, 1+\sqrt{-5}>.
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  3. #3
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    there´s one thing that´s not clear to me:
    I´ve just proved that \ker f = <3,1+\sqrt{-5}>, which means that \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}> is isomorphic to \mathbb{Z}3, which is a field.
    That allows me to say that the ideal is prime?
    As far as I´m concerned, the quotient of a ring by a prime ideal is a domain.
    I´m kind of confused with that...
    But the help was very helpful


    And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??

    Thanks!
    Last edited by Inti; January 16th 2009 at 05:13 AM.
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  4. #4
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    Quote Originally Posted by Inti View Post

    there´s one thing that´s not clear to me:
    I´ve just proved that \ker f = <3,1+\sqrt{-5}>, which means that \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}> is isomorphic to \mathbb{Z}3, which is a field.
    That allows me to say that the ideal is prime?
    that means the ideal is maximal and we know that every maximal ideal is prime.



    And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??
    no. if I and J are two ideals, then IJ=\{\sum_{k=1}^na_kb_k: \ n \in \mathbb{N}, \ a_k \in I, b_k \in J \}. however, for example, if you want to prove that IJ \subseteq K, where K is an ideal, then you only need to prove that

    ab \in K for all a \in I, b \in J.
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